calculus

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A point moving on a coordinate line has position function s. Find the velocity and acceleration at the t, and describe the motion of the point during the indicated time interval

s(t)=-2t^3+15t^2-24t-6 [0,5]

  • calculus -

    v = ds/dt = -6 t^2 + 30 t -24

    a = dv/dt = -12 t + 30

    at t = 0 the point is at -6
    at t = 1 it is at -17
    at t = 2 it is at -10
    etc until
    at t = 5 it is at -1
    graph that
    the velocity changed sign in there, look for where v = 0
    -6 t^2 +30 t -24 = 0
    -t^2 + 5 t - 4 = 0
    t^2 -5 t + 4 = 0
    (t -4)(t-1) = 0
    at t = 1 and at t = 4 the direction reverses
    at t = 0 the velocity is -24 so it starts out at -6 moving south fast
    at t = 1 it reverses course and starts up from -17 etc
    at t = 4 it is at 10 and starts down toward reaching 1 at t = -5

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